Question

If the wavelength of the first line of the Lyman series for the hydrogen atom is $1216{{A}^{\circ }}$, then the wavelength of the first line of the Balmer series of the hydrogen spectrum is:
$\begin{align}
  & (A)1216{{A}^{\circ }} \\
 & (B)6563{{A}^{\circ }} \\
 & (C)912{{A}^{\circ }} \\
 & (D)3648{{A}^{\circ }} \\
\end{align}$

Answer 1

Hint: In Lyman series the transition occurs from the first level to higher levels whereas in Balmer series the transition occurs from the second level to the higher levels. The wavelength of the Lyman series is given in the equation hence using the Rydberg formula calculates the wavelength of the Balmer series.
Formula used:
The Rydberg formula is given by,
\[\] $\dfrac{1}{\lambda }=R\left[ \dfrac{1}{n_{1}^{2}}-\dfrac{1}{n_{2}^{2}} \right]$
Where R is the Rydberg constant.
\[\lambda \]is the wavelength of the spectral line in the hydrogen spectrum.
${{n}_{1}}\And {{n}_{2}}$are the levels of transitions.

Complete step by step answer:
In the case of the lyman series the transition of electrons takes place from the first level to the higher levels. Hence Rydberg formula takes the form,
$\dfrac{1}{{{\lambda }_{L}}}=R\left[ \dfrac{1}{n_{1}^{2}}-\dfrac{1}{n_{2}^{2}} \right]$
$\dfrac{1}{{{\lambda }_{L}}}=R\left[ \dfrac{1}{1_{{}}^{2}}-\dfrac{1}{{{2}^{2}}} \right]$
$\begin{align}
  & \Rightarrow \dfrac{1}{{{\lambda }_{L}}}=R\left[ \dfrac{1}{1}-\dfrac{1}{4} \right] \\
 & \\
\end{align}$
$\begin{align}
  & ie,\dfrac{1}{{{\lambda }_{L}}}=R\left[ \dfrac{3}{4} \right] \\
 & \therefore {{\lambda }_{L}}=\dfrac{4}{3R} \\
\end{align}$
Given that,
${{\lambda }_{L}}=1216{{A}^{\circ }}$
Similarly, in the case of the Balmer series the transition occurs from the second state to the higher states. That is,
$\dfrac{1}{{{\lambda }_{B}}}=R\left[ \dfrac{1}{n_{2}^{2}}-\dfrac{1}{n_{3}^{2}} \right]$
\[\]$\begin{align}
  & \dfrac{1}{{{\lambda }_{B}}}=R\left[ \dfrac{1}{{{2}^{2}}}-\dfrac{1}{{{3}^{2}}} \right] \\
 & \dfrac{1}{{{\lambda }_{B}}}=R\left[ \dfrac{5}{36} \right] \\
\end{align}$
${{\lambda }_{B}}=\dfrac{36}{5R}$
Taking the ratio of wavelength of Balmer series to the wavelength of the Lyman series the equation becomes,
\[\begin{align}
  & \dfrac{{{\lambda }_{B}}}{{{\lambda }_{L}}}=\dfrac{\dfrac{36}{5R}}{\dfrac{4}{3R}} \\
 & \dfrac{{{\lambda }_{B}}}{{{\lambda }_{L}}}=\dfrac{27}{5} \\
 & \therefore {{\lambda }_{B}}=\dfrac{27}{5}{{\lambda }_{_{_{L}}}} \\
\end{align}\]
Hence substituting the value of wavelength of lyman series in the above equation we get,
${{\lambda }_{B}}=\dfrac{27}{5}\times 1216$
${{\lambda }_{B}}=6563{{A}^{\circ }}$

So, the correct answer is “Option B”.

Note: Lyman series is visible in the ultraviolet region of the hydrogen spectral series. The Balmer series is seen in the visible region of the hydrogen spectrum whereas Paschen and pfund series are seen in the infrared region of the hydrogen spectral series.